It is the Example 1.10 from the book Introduction to probability Models.

Let a ball drawn from an urn containing four balls, numbered 1,2,3,4. Let E={1,2},F={1,3},G={1,4}. If all four outcomes are assumed equally likely, then

P(EF) = P(E)P(F) = 1/4 
P(EG) = P(E)P(G) = 1/4
P(FG) = P(F)P(G) = 1/4


1/4 = P(EFG) not equal to P(E)P(F)P(G)

Hence, even though the events E,F,G are pairwise independent, they are not jointly independent.

My question:

  1. Does P(EFG),P(EF),P(E) or P(F) have an intuitive meaning here respectively ? What is sample space for P(EF),P(E),P(EFG)?

I doubt P(EF) means the probability of draw 2 balls from the urn and one of them is {1}, P(EFG) means draw 3 ball from the urn and one of them is {1}?

  1. How to calculate P(E) or P(F) here? Is their value equal to 1/2 respectively?

After perfect comments made by @JMoravitz.

  1. Regarding independent events, can I understand P(EF) in this way intuitively? To get a {1} within the scope of event E or F. Event E and event F are equally likely, and in event E or event F, {1} has 1/2 chance to be picked. so that P(EF) = 1/2(E or F) * 1/2.
  • 2
    $\begingroup$ By $EF$ they mean $E\cap F$, that is the intersection of the events. $E$ occurring and $F$ occurring simultaneously. You still draw only a single ball. In words, you might phrase it $E$ is the event that you draw a small number (small in this case meaning a 1 or a 2) and $F$ is the event that you draw an odd number. $E\cap F$ is the event that you draw a small odd number (i.e. you draw a one). Similarly, $EFG$ here is shorthand for $E\cap F\cap G$ is the event that you draw a small odd number which is one of the outside numbers (i.e. you draw a one). $\endgroup$ – JMoravitz Aug 26 '16 at 5:08
  • 1
    $\begingroup$ As for calculating $Pr(E)$, note that $E=\{1,2\}=\{1\}\cup\{2\}$. We are told that $Pr(\{1\})=Pr(\{2\})=Pr(\{3\})=Pr(\{4\})$ and we know that our sample space is $\Omega=\{1,2,3,4\}$, implying $1= Pr(\Omega)=Pr(\{1\}\cup\{2\}\cup\{3\}\cup \{4\})=Pr(\{1\}) +\dots+Pr(\{4\})=4Pr(\{1\})$ further implying that $Pr(\{1\})=\frac{1}{4}$. So we have $Pr(E)=Pr(\{1\}\cup\{2\})=Pr(\{1\})+Pr(\{2\})-Pr(\{1\}\cap\{2\})=\frac{1}{4}+\frac{1}{4}-0=\frac{2}{4}=\frac{1}{2}$. Similarly for $F$ $\endgroup$ – JMoravitz Aug 26 '16 at 5:11
  • $\begingroup$ @JMoravitz These comments answer the question and, as such, should really be posted as an answer not a comment. (Unless you're already typing up something right now, and this is just a place holder.) $\endgroup$ – Graham Kemp Aug 26 '16 at 5:34
  • $\begingroup$ @JMoravitz, Thanks for your detailed explanation. I misunderstood the event space. Yes, I can understand the meanings of P(E),P(F),P(EF),P(EFG) by the Set theory. $\endgroup$ – evergreenhomeland Aug 26 '16 at 6:19

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